求助，问个力扣上的题，有个bug请大佬帮忙看看

哥链接给我我来做做

你确定没有进if ？
但凡进了if时，i不是1，cnt就不会是-1，就无法break

有木有可能，cnt=0而不是-1呢！对于不确定的，不妨加个printf

这题不是判断最后一个数字是不是0就可以了吗
在额外考虑一下本身就是0的问题

#include <stdlib.h>
#include <math.h>
#include <memory.h>
#include <iostream>
class Solution {
public:
bool isSameAfterReversals(int num);
};
constexpr int N = 64;
const char *arr = "0123456789abcdefghigklmnopqrstuvwxyz";
const char *w[2] = {"false", "true"};
char **p = (char **)w;
char *itoa(long long num, char *buf, int d)
{
memset(buf, '\0', N * sizeof(char));
char *p, c;
long long b = 0, h = num, i = 0;
while(h > 0)
{
b = (long long)fmod(h, d);
c = *(arr + b);
buf[i++] = c;
h /= d;
}
for (long long j = strlen(buf) - 1, i = 0; i <= j; i++, j--) //逆转数组
{
c = buf[j];
buf[j] = buf[i];
buf[i] = c;
}
p = buf;
return p;
}
long long atoi(char *str){
long long p = 0, f = 0, r = (long long)strlen(str);
if(r < 0)
{
return -1;
}
for(long long i = 0; i < r; i++)
{
f = (int)(str[i] - '0');
if(f == 0)
{
continue;
}
p += f * powl(10, r - i - 1);
}
return p;
}
bool Solution::isSameAfterReversals(int num) {
if(num > INT_MAX)
{
return false;
}
char buf[N];
itoa(num, buf, 10);
char t[N];
memset(t, '\0', N * sizeof(char));
int g = strlen(buf);
for(int d = 0; d < g; d++) {
t[d] = buf[g - d - 1];
}
long long s = atoi(t);
itoa(s, buf, 10);
g = strlen(buf);
memset(t, '\0', N * sizeof(char));
for(int d = 0; d < g; d++)
{
t[d] = buf[g - d - 1];
}
long long j = atoi(t), st = (long long)num;
bool ut = (j == st);
std::cout << w[(int)ut] << std::endl;
return ut;
}

这道题判断一下是不是零和最后一个数字是不是零就够了啊。还有为什么设置这么多数组你这一百万个数组大小都不够了吧